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Reading aterial
for
asic Electricity
Reading Material for Basic Electricity
Electrical
Electronics Laboratory
Centre
for H R D
Durgapur Steel Plant
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BASIC ELECTRICITY
CONTENT
SI.
Topics
Page No.
1
Source of E lectricity 1
2 Ohm's Law
1
3 Laws of Resistance,Resistance in series and Uses of Series Resistance.
2 -3
4
Resistance in Pa rallel and U ses of parallel Resistance.
3 - 4
5
Principle of Inductor, Behavier of L-R Ckt with DC
4
6
Behavier of R - C Ckt with DC
5
7
Charging & Discharging of Capacitor with DC
6
8 Di- Electic Strenth of Capacitor
7
9
Capacitor in Series & P arallel
7
10
Defination & N ature of AC Volages
8
11
RM S V alue of an A C V oltage / Cu rrent, Electromagne tic Induction
9
12
Resistance & C apacitor Colour Code chart 10 - 12
13
Practical Assignment on Resistive Ckt at AC / DC
13 - 15
14
Practical Assignment on Inductive Ckt at AC
15- 16
15
Practical Assignment on Capcitive Reactance ( X c)
16 - 17
16 Practical Assignment on Inductive Reactance ( X L)
17 - 18
17
Practical Assignment on Series RC ckt.
18- 19
18 Practical Assignment onS eries LR Ckt.
19
19
Practical Assignment on impe dence of a Series CR Ckt & app lication.
20 -22
20
Practical Assignme nt on impede nce of a Series LR Ckt & ap plication.
23 - 24
21
Practical Assignment on impedence of a Series LCR Ckt & application.
25 - 26
22
Practical Assignment on Parallel and app lication
27--29
23
Working Principle of a Transformer, E.M.F. Equation
29 - 30
24
Power at AC ( 1 Ph & 3 Ph)
30-31
25
Meggering
31- 33
26
Safety Precautions for Electricians.
34-35
27
Treatme nt of Electical shock for Artificial Respiration
36 -
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Source of E lectricity
Electrical energy m ay be produced in several ways. Basically there are five m ethods
of
providing voltage and current.
Friction
This m ethod allows for the separation of electrons and protons b y
the work
of rubbing tw o dissimilar materials together wh ich, in turn produces static
electricity
because of the separation an d storage of opposite charges.
Chemical energy
this metho d is generally found in dry cells or storage
batteries.
Fund am entally, it involes a chem ical reaction between two dissim ilar metals and
electrolyte in w hich charges of o pposite polarities are produced. The m etals
which
beco m es deficient in electrons bec om es the positive plate, and the m etal with ;:r1
abund ance of electrons becom es the negative plate. The chem ical action
continues
until no further transfer of electrons is possible.
Photoelectricity
thou ght are use of the ph otoelectricity effect, light energy
can t
.4
,
;
con verted into electrical energy . The process involes coating the surfaces
of
m etals w ith pho tosensit ive m aterials such as selenium or caesium .
Where
strikes the m aterials, P
hotoelectrons are em itted and create a current flow
by tho
,
steady movement.
Thermal Energy
as the nam e indicates thermal energy involves the
convers:cn
heat into electrical energy. This can be accom plished by h eating metals
which
turn supp ly energy to the free electrons. The electrons in turn break aw ay
from Olt.:
atoms,
creating a current.
Magnestism
M otion can b e applied to a conductor wh ich, when allowed to
move
a magnetic field create a current. This is basically the principle of
generaro
operation.
OHM'S LAW
Temperature and other physical conditions rem aining constant, the potential
different
across the two terminals of a cu rrent carrying con ductor is directly proportional to
the
current flowing through it.
Frv A B is a conductor and the potential at points A & B are VA and Vg
;.pose VA > VE, So potential difference is VA ..V3 = Say V )
Let us assume that when the conductor is connected to battery B (of emf = V),
current
( I
flows through the condu ctor.
According to Ohms law ; A VB ) cc I
or, V I or V = IR
As VA VB = V )
wh ere R is the resistance of the conductor and its unit is ohm Un it of voltage and that of
I s Am p.
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V
I V
Law s of resistance
R oc p L / A
w here, P = Speci f ic resistance, L= Len gth of the condu ctor,
A = Cross sectional area of the con ductor.
Resistance is directly proportional to the length of the con ductor.
Resistance is inversely proportional to the area of cross section of the
conductor.
Value of the specific.resistance of the cond uctor depend s upon the properties
of conductor m aterial and the tem perature of the condu ctor. Thus a nichrome
wire
Has about Sixty six times more resistance than a copper wire of the same
dimen sions at sam e temperature.
RESISTANCE IN SERIES
R2
Notice that the sum voltage across each resistor is equal to the applied voltage
(V) i.e. total voltage. V = Vi+ V2+V3
Accord ing to Ohm 's law th is can be expanded as IR=IRi + IR2 + IR 3 A s
resistors are connected in series, the sam e value of current (say I ) must flow in
the circuit. So R1 = R1 +R2+ R 3 This Shows that for resistors in series, the
equivalent resistance is equ al to the sum of the resistors in series.
This formula can be writ ten as: R1 = R1 +R2 +R3
Rn for any num bers of
resistances in se ries.
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Capacitor
C
o
ka.5e
-t.-t.rir 7 C
741-1c.
Behavior of R-- C Circuit with DC.
In our earlier assignment we have known about condenser or capacitors. Now for
knowing its nature with DC voltage we shall see the following figure. Here the action wiii
opposite to that in the circuit using an inductor. The initial current will be and the current
will start decreasing until the current become zero and capacitor gets fully charged . The
current curve will be like the following figure.
R
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414_
Charging
ischarging
CAPACITOR
Capacitor can be defined as two conductors separated by an insulating dielectric. It is a
device to store electrical energy and to release it when required. In figure a simple
parallel plate capacitor is connected with a battery.
Battery
C
CHARGING OF CAPACITOR
Working Principal :-- Suppose plate A is connected to +ve
pole and plate B to -ye pole
of DC supply. On closing the switch there will be momentary flow of electrons in the
direction indicated some electrons are withdrawn from the plate A leaving it positively
charged and transferred to plate B giving it a negative charge. This flow of electrons
gives current which decreases and finally ceases when the voltage across the capacitor
plate has become equal and opposite the applied voltage. A capacitor then blocks DC
circuit . The charged capacitor is now a store of energy. The practical units
of capacity
is farad. A conductor is said to possess capacity of one farad if its potential is raised by
one volt when one coulomb of charge is given to it.
One farad = 1 Coulomb / 1 voltage
Sub-units of Farad are micro and pico farads.
1p F =1 MF = 1/ 1000000 farad
1pp F or 1 pico farad = 1/ 1000000000000
Factors on which the capacity of capacitor depends :
The area of the plate of the condenser (directly proportional).
The distance between the plate ( inversely proportional).
The nature of dielectric, i.e. dielectric constant.
[ C = K* A/d where A = Area, D = distance , K = dielectric constant ].
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V3
2
voltage
Di- Electric Strength
The maximum kilovolts per millimeter or Volts per mm which a medium can withstand
without break-down, is known as its
Di-electric
strength .
The insulating materiel which resists the break-down is called its dielectric strength e.g.'
Glass = 200 -250 KV / mm
Mica = 500-1500 KV / mm .
The various factors on which the dielectric strength depends are :
Thickness' of material. Break-down voltage (E) is not proportional to the thickness
(T). E =
A* t
2
/
3
) where A is materiel constant .
Temperture. It decreases with rise of temperature .
Absorbed moisture content affects inversely .
Shape and size of electrodes.
Capacitor in series
Let C1, C2, C3 = Capacitance of the capacitors connected in series (figure )
V1, V2, V3 = P. D. drop of
three Capacitors
respectively. ,
V =The total applied
C = Joint capacity
Now in series, V = V1 + V2+ V3
or Q/C = Q/C1+Q/C2+Q/C3 ( since V = Q/C in condenser)
or 1/C = 1/C
1
+1/C2+1/C3 (Taking Q as common ) .
In series combination, charge of all capacitors is the same, but P.D. across each is
different. In this case the reciprocal of the total capacitance is equal to the sum of the
reciprocals of individual capacitance's.
Capacitors in parallel
The three capacitors C1,C2 and C3 are connected in parallel to supply voltage V.
Let the charged on the capacitors be Q1, Q2,& Q3.
Then Q1 = C1 V ,
2 = C2 V,
3 = C3 V
Let C = Equivalent capacitance, then Q = Q1+Q2+Q3
therefore C = C1+C2+C3 .Thus the total capacity is equal to the sum of the
individual-capacitance's .
ci
2
3
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Vmax
Vp-p
Triangular wave
Square wave
DEFINITION AND NATURE OF A.0 VOLTAGES
So far we have default with pure D.0 voltage and that too only constant magnitude . But
there are other voltages too e.g. varying D.C. voltages of various shapes. Now let us
look at the various wave from below to understand what A.C. voltage is,
The polarity of an A.G. voltage always changes after the completion of every half cycle
and after every full cycle the full wave shape repeats it self . In case of pure A.C. wave
from the area of 'positive' and 'negative' half cycles are equal. Then. of full cycles which
occur in one second time is called frequency of the A.C. voltage. In fact, all voltage
generated out of rotating machines are basically A.G. voltage. But in case of D.C.
generators it is the commutator that converts the A.C. to D.C. voltage. Moreover voltage
level changing in case of A.G. can be done very easily. So we must have a knowledge
of A.G. voltages.
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R.M.S. VALUE OF AN A.C. VOLTAGE / CURRENT
As the instantaneous values of any AC. waveform are always changing, when we
measure the magnitude of an A.C. voltage by an AVO meter we actually get its R.M.S.
value ( Root Mean Square value). The R.M.S. value of an alternating current is giving
by that steady ( D.C.) current which when flowing through a the same circuit for the
same time.
For example, RMS ( for a sine wave ) = V peak /
or RMS ( for a sine wave) = I peak / 42
ELECTRO-MAGNETIC INDUCTION.
Whenever there is a change of flux linked with any circuit ,an emf, is induced in
the circuit. This is known as Faraday's 1st law of induction. The amount of emf enduced
will depend on the time rate of change of flux linkage . This refers to the 2nd law of
Faraday. The direction of induced emf will always oppose the cause producing the
induced emf. This is called Lenz's law.
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R CAPACITOR INDENTIFICATION CHART
Colour
Band
A
Band
B
Band C
and D
( Multiplier ) Tolerance )
Band E
Resistor
Capacitor
Resistor
1%
0pF
Capacitors
Resis-
for
---
---
20%
Poly-
-ester
Capa-
-citor
---
---
Upto
2pF
0.1pF
Over
10pF
1%
Black
---
0
10
10
---1
Brown
1
1
10'
1F
-
10
1
10
-2-
10
3
Red
2
2
2%
---
---
---
2%
2.5%
250w
---
---
---
range
3
3
103
Yellow
4
4
10
0
---
---
---
---
Green
5
5
10
5
---
---
---
0.5pF
---
5%
---
_ _ _
---
---
---
lue
6
6
10
6
---
Violet
7
7
107
---
---
---
---
---
---
Grey
8
8
108
0.01pF
---
0.25pF
--- ---
---
White
9
9
10
9
01 pF
---
10%
1pF
---
10%
---
---
---
---
---
ilver
---
---
0.01
---
Gold ---
---
0.1
---
5%
---
---
--- ---
Pink
---
--- ---
---
--- ---
---
---
Hi-
Stab
None
---
---
---
---
20%
---
---
---
Resistor Capacitor letter And Digit code ( BS1852)
Preferred value
E 12
Series
1.0 1.2
1.5 1.8
2.2 2.7
3.3 3.9 4.7 5.6
6.8
8.2 and their
decades.
E 24
Series
1.0
1.1 1.2
1.3 1.5
1.6 1.8 2.0
2.2
2.4 2.7 3.0
.3
3.6
3.9 4.3 4.7 5.1 5.6 6.2
6.8 7.5
8.2
9.1
and their
decades.
Resistor Capacitor letter And Digit code ( BS1852)
Resistor values are indicated as follows:
0.4752
marked
R47
152
marked
1R0
4.70
Marked
4R7
47Q
Marked
47R
100Q
Marked
100R
1k0
Marked
1K0
10k0
Marked
10K
10MQ
marked
10M
A letter following the value shows the tolerance. F =1%: G =2% : J = 5% :
K = 10% : M = 20% : R33M = 0.33Q20% : 6KSF = 6.81(Q1%
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Capacitor values are indicated as follows
:
0.68 pF
Marked
p68
6.8 of
Marked
6n8
6.80 pf
Marked
6p8
1000 nF
Marked
1p0
1000 pF '
Marked 1n0 _ _ .8 pF
Marked
6p8
Tolerance is indicated by letter as for resistors. Values up to 999 pF are marked in pF
from 1000pf.to999000pF = 999nF as nF 1000pF = 1nF and from 1000nF( = pF
upwards as pF.
Some capacitors are marked with a code denoting the value in pF (first two
figures)followed by a multiplier as a power of ten (3 =10 ).Letters denote tolerance as for
resistors but C = 0.25pf. E.g.123J = 12pF x 10
3
5% =12000pF (or 0.12pF).
Tantanum Capacitors
Color 1
2
3
4
Color
1
2 3 4
Black
---- 0 x1 10V
Blue
6
6 ----
20
Brown 1 1 x10 ---- Violet
7 7
----
----
Red 2 2
x100
----
Grey
8
8 x0.01 25V
Orange 3
3
----
---- White 9
9 X0.1 3V
Yellow 4 4
----
6.3V
Pink
10
10 ---- 35V
Green
5 5
---- 16V
Reactance of Capacitors at spot frequencies.
50Hz
100Hz
1kHz 10kHz 100kHz 1MHz 10MHz 100MHz
1pF
----
----
----
---- 1.6M
160k
16k
1.6k
10pF
---- ----
---- 1.6M
160k
16k 1.6k 160
50pF
----
----
3.2M 320k 32k
3.2k
320 32
250pF ----
6.4Mm
640k
64k
6.4k
640
64 6.4
1000pF 3.2M
1.6M
160k
16k 1.6k
160 16 1.6
2000pF 1.6M 800k
80k
8k 800
80 8 0.8
0.01pF 320k 160k 16k
1.6k 160 16
1.6 0.16
0.06pF 64k
32k
3.2k
320
32
3.2
0.32
----
0.1pF
32k
16k
1.6k 160 16 1.6
0.16 ----
1pF 32k 1.6k 160 16
1.6
0.16
----
----
2.5pF 1.3k 640
64
6.4
0.64 ---- ---- ----
5pF 640 320
32
3.2 0.32 ---- ---- ----
10pF
320 160
16 1.6 0.16
---- ---- ----
30pF
107 53
5.3
0.53 ---- ---- ---- ----
100pF 32
16 1.6
0.16 ---- ---- ---- ----
1000pF
3.2
1.6
0.16
----
---- ---- ----
----
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Reactance of Inductors at spot frequencies
50Hz
100Hz
1kHz
10kHz 100kHz
1MHz 10MHZ 100MHz
1pH ----
----
----
----
0.63 6.3 63 630
5pH ---- ---- ---- 0.31 3.1
31
310
3.1k
10pH
----
----
----
0.63
6.3
63 630
6.3k
50pH
----
----
0.31
3.1
31
310
3.1k
31k
100pH
----
----
0.63
6.3
63
630
6.3k 63k
250pH
0.16
1.6
16
160
1.6k 16k 160k
1mH
0.31 0.63
6.3 63 630
6.3k 63k 630k
2.5mH
0.8
1.6
16 160
1.6k 16k
160k 1.6M
10mH
3.1
6.3
63
630
6.3k 63k
630k
6.3M
25mH 8
16
160 1.6k
16k
160k
1.6M ----
100mH 31
63 630
6.3k
63k
630k 6.3M ----
1H
310
630
6.3k
63k
630k
6.3M ----
----
5H
1.5k 3.1k
31
310k
3.1M
----
----
----
10H
3.1k 6.3k 63k 630k 6.3M
----
----
----
100H
31k 63k 630k 63M
----
---- ---- ----
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ASSIGNM ENT ON RE SIST IVE CU RCU IT AT AC / DC
OBJECTIVE To investigate a resistive circuit at AC
PROCEDURE
Connect the power supply unit to the mains supply line. DO NOT
switch on yet.
EXPERIMENTAL PROCEDURE
We wish now to see how a resistor behave at a. c.
Connect up the circuit as shown in the patching diagram of figure.
Ensure the variable d. c. vo cage control knob is turned fully counterclockwise, then
switch on the pus. Very the d.c control slowly, and observe the two meters.
PRACTICAL CONSIDERATIONS AND APPLICATIONS Resistors are used at a. c. in
similar ways to those d.c applications of earlier assignments. There is no phase shift
across a true resistance, and the ratio of voltage to current shown by a resistance is
constant with frequency, so the behavior at a/c is no different from its behavior at d.c.
When calculating power dissipation at a.c, rms. values of voltage and current must
apply.
PRACTICAL CONCIDERATIONS AND APPLICATIONS Resistors are used at a.c
in similar ways to those d.c applications of earlier assignments . There is no phase
shift across a true resistance , and the ratio of voltage to current shown by a resistance
is constant with frequency , so the behaviour at a.c is no different from its behaviour at
d.c. When calculating power dissipation at A.G., rms values of voltage and current must
apply.
Practical Assignment on CAPACITIVE ckt. at AC
OBJECTIVE
To investigate a capacitive circuit at a. c.
PRELIMINARY PROCEDURE
We known, from assignment 8, that the relationship
between charge, voltage and capacitance is : Q = CV also we known :Q = It, where I is
current and t is time. From these we can say that if a capacitor of C farad is charged
from OV to V volts, in t seconds then : charging current, I= charging current = capacitor
x rate of increase of voltage. Let us see what happens when a sinusoidal alternating
voltage is applied to a capacitor. Connect up the circuit as shown in the patching
diagram of figure , corresponding to the circuit diagram of figure
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2.2
micro
F
AC
CH1
CH2
100 R
orn
Set the function generator to give a 10Vpeak-to-peak sine waveform at 250Hz.Set the
Oscilloscope as follows : Y1 channel 1V/cm. Y2 channel 500mV/cm .Time base to
1ms/cm. Zero both the traces and then observe the two waveforms on the oscilloscope.
Mathematically, if the voltage waveform is denoted by the formula :
V = V
max sin t
then as i = C x rate of change of voltage
I = C / dv /dt = C VmAx d /dt (sin w t)
I =CV Ax cos w t
Thus if v is sinusoidal , i.e the same shape , but leading by 900 .
This is because cos t =sin( t+90
)
If we were to plot the voltage and current waveforms in the capacitor by that method we
would require two vectors . Both vectors would rotate while keeping a constant 90
0
angle between them , as shown in figure. As we go on we shall find it useful to think in
terms of these vector, but rather confusing if they are always rotating. Usually it is the
relationships between them that are important, as for instance the 90
0
angle between
those in figure .These relationships can be studied conveniently in a
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CH
CH2
orn
Inductor
T
Resistor
FG
diagram where the vectors are shown at rest. The vectors are then said to be
represented by 'phasor' and the diagram is a 'phasor diagram'. Figure is phasor
diagram corresponding to figure.
The voltage phasor is taken as the reference and is drawn horizontally pointing to the
right (3 'o'clock). The current phasor leads the voltage phasor by 900
and is thus drawn
900 counterclockwise from the reference. When an alternating voltage is applied across
a capacitor an alternating current flows. Yet when a.d.c voltage is applied, after an initial
flow of charging current, no d.c current flows. This behaviour is different from that of a
resistance. Never the less if an a.c voltage and an a.c current can exist, the ratio
between them is likely to be interest, and the ratio is therefore given a different name. In
an a.c circuit the ratio of voltage to current is called 'impedance' and is denoted by Z.
Thus in an a.c circuit Z = We shall examine this idea further in other assignment. For
the moment it may be noted that impedance may be taken as the ratio of two phasor,
and therefore has both magnitude and phase. Magnitude = Z = Phase
of the
impedance is the angle between the phasor. The impedance of a capacitor has a phase
of 90 radians Impedance's of 900
phase angle have special properties and
are
given the special name 'reactance'.
PRACTICAL CONSIDERATIONS AND APPLICATIONS
The applications
section of
Assignment gives details of capacitors in general, but there are a few more
points
which
appear
at high frequencies(hf).Due to
the form
of construction' wound capacitors of any
type possess appreciable inductance, thus do not act as pure capacitors. The effects of
this inductance are greater at high frequencies and thus wound from of capacitors is
inferior at h.f. Mica capacitors, ceramic tubular and disc capacitors, and air dielectric
capanitnrs
are
ideal
fnr high frequency
work
Capacitors normally have a stated
maximum voltage rating above which they cannot be used safely. When used at a.c the
peak or crest value of the voltage must be used to determine wher ther the capacitor is
within rating.
PRACTICAL ON ASSIGNMENT INDUCTIVE CIRCUIT AT AC
OBJECTIVE
To investigate an inductive circuit at a.c.
EXPERIMENTAL PROCEDURE From assignment 12 we have found
out
that the
relationship between induced emf, current and inductance in a system is : e = - L or :
(Induced emf ) = - (inductance)(rate of change of current) We will now examine what
happens when a sinusoidal alternating voltage is applied to an inductor. Connect up the
circuit as shown in the patching diagram of figure 16.2, corresponding to the circuit
diagram of figure
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Set the function generator to give a 10V peak-to-peak waveform at 250Hz. Set the
oscilloscope as follows : Y1 channel(current) to 1V/cm Y2 channel(voltage) to50OrnV
/cm Time base to 1ms/cm. Zero both traces, the observe the two wave forms on the
oscilloscope. Draw the two waveforms you see, showing their relative positions with
respect to each other.
PRACTIAL CONSIDERTIONS AND APPLICATIONS
By its nature of operation an
inductance will create a magnetic field around it. With an alternating current flowing
through the inductor the magnetic field around it will be alternating. If the inductor is the
presence of other components or conductors this alternating magnetic field will link
with these conductors and induce emfs in them. These emfs will generally be unwanted,
and give rise to noise, hum, or interfering signals.. For this reason, inductors are often
magnetically screened in cans, or housings made from a non - magnetic material such
as mumetal or aluminium. When designing inductors care must be taken in the selection
of type and gauge of wire used. Obviously the lowest possible resistance is desired and
at low frequencies this means the thickest possible wire gauge consistent with a
reasonable sized winding, however at high frequencies multi-stranded litz wire is often
better to minimise the skin effect.
ASSIGNMENT on CAPACITIVE REACTANCE ( Xc )
OBJECTIVE
To investigate the impedance of a capacitor to an a.c sinusoidal waveform
and see how this varies with frequency.
EXPERIMENTAL PROCEDURE
In a capacitor a sinusoidal current and voltage are
always 90
0
out of phase with one another , so that the phase of the impedance is
constant .The magnitude varies however, in a manner which we shall now discover.
connect up the circuit as shown in the patching diagram of figure . corresponding
to the
circuit diagram of figure.
C
Copy the results table as shown in figure, reproduced at the end of this assignment. Set
the frequency of the function generator to 800Hz. Adjust the output of the generator to
give 1V rms as read on the matter. Take the reading for this voltage. Reset the voltage
output of the generator to 2V rms. Record the resultant current. Repeat this procedure
for voltage of 3V, 4V, 5V and 6V rms. Record your results and calculate the ratio of rms
voltage to rms current.
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PRACTICAL CONSIDERATIONS AND APPLICATIONS
capacitors show an infinite
impedance to d.c but their impedance is infinite to a.c and decreases as the frequency
of the a.c increases. Therefore one of their used is in coupling between circuits or parts
of circuit, where it is wanted to block any d.c transmission but to allow a.c signals to
pass. Large value electrolytic capacitors are often used in 'smoothing' circuits in which a
d.c voltage which has an a.c ripple voltage superimposed upon it is applied to the
capacitor. The capacitor has no the effect on d.c component of the wave form, as its
impedance to d.c is infinite, but by passes to earth the a.c. This is shown in figure .
PRACTICAL ASSIGNMENT ON INDUCTIVE REACTANCE ( XL )
OBJECTIVE
To investigate the impedance of an inductor to a sinusoidal a.c
Waveform .
E XP E R IME N T AL P R O C E D UR E
We have said that impedance , Z is given by
Z = V
rms rms
And that , for a capacitor , its impedance termed capacitive reactance,. X c .
Similarly we can determine the INDUCTIVE REACTANCE of an inductor. Inductive
reactive is given the symbol XL .Connect up the circuit as shown on the patching
diagram of figure 18.2, corresponding to the circuit diagram of figure 18.1
L
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R3
Copy the results table as shown in figure reproduced at the end of this assignment.
Adjust the generator to give a frequency of 4kHz and an output of 1V rms sine wave, as
shown on the mater. Record the current reading for this voltage. Readjust the output
amplitude to 2V rms and record the resultant current. Repeat this for voltage steps of 3V
4V ,5V and 6V rms. Record your results and calculate the impedance for each step.
PRACTICAL CONSIDERATIONS AND APPLCATIONS
the inductor is the converse
of the capacitor in that it shows an impedance which is practically zero
to d.c but
increases with frequency .Like a capacitor , an inductor can also be used in smoothing
circuits, but it is used in series with the d.c line instead of across it. See figure . The d.c.
is passed without effect while the a.c is greatly impeded by the inductor. In this
application the inductor is often called a 'choke'.
L
t
v i e . .
When an inductor contains magnetic material such as iron , it can only maintain its
inductance over a limited range of current . Excessive current cause the 'iron' to
saturate (i.e
fail to permit more than a limited amount of flux). The inductance for
therefore decreases. This is why you were advised to omit any reading for which the
current exceeded 60mA.
ASSIGNMENT THE SERIES C R CIRCUIT
OBJECTIVE
To investigate the series CR circuit and determine the relationships
governing amplitude and phase shift .
EXPERIMENTAL PROCEDURE
We know that voltage and current are in phase in a
resistive circuit , and that , for a capacitive circuit , the current leads the
voltage by 90
Lets us investigate what happens in a circuit consisting of both capacitance and
resistance . Connect up the circuit as shown in the patching diagram of figure
corresponding to the circuit diagram of figure .
V
.....
......... ......
....
.......
.........
0
C
l
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Set the Fun ct ion ge nerator f requen cy to 50Hz , and i ts output to 20V p eak-to-peak ,
Sine wave as sho wn on the Y1 chan nel of the oscil loscope .
Have the o scilloscope set to : Y1 chann el(Vi
n
) to 5V / cm . Y2 channel ( ) to 5V /cm
Time b ase to 5mS / cm . Zero both the oscilloscope traces .
W ith link 1 connec ted as show n in f igure 19.2 , ie by pass ing the resistor chain so that
resistor is zero , observe the waveform for and V
c . They should be superimp osed .
M easure their amp litude . Now conn ected l ink 1 betw een point
A and D to leave the
1 kS1 resistor in series with the capacitor .Measu re the am plitude o f V
c
and the phase
shift m ay be m easured as fol lows :
ASSIGNMENT THE SERIES LR CIRCUIT
OB JECTIVE To invest igate the series LR circuit and determ ine the relationships
governing am plitude and phase sh ift .
EXPER IMEN TAL PR OC EDU RE W e have investigate wh at happens in a series RC
circuit when R is varied , let us now see wh at happens w hen R is varied in an LR circuit .
Con nect up the circuit as show n in the patching diagram of figure
corresponding to the
circuit diagram of figure .
0--
R 1
2
--o-H
*-1
R.3
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0
PRACTICAL ASSIGNMENT IMPEDANCE OF A SERIES CR CIRCUIT
OBJECTIVE
To investigate the series CR circuit and determine the relationships
governing amplitude and phase shift .
EXPERIMENTAL PROCEDURE
We have investigate the impedance ( or reactance ) of
a purely capacitive and an inductive circuit , and arrived at formulae relating impedance
with frequency and capacitance or inductance . Let us now see if we can arrive at some
formulae for the series CR circuit of figure . Connect up the circuit as shown in the
patching diagram of figure corresponding to the circuit diagram of figure .
Set the frequency of the function generator to 100Hz Adjust the
output of the generator
to give 1V rms as read on the meter . Take the current reading for this voltage .
Reset
the voltage output of the generator to 2V rms .Record the resultant current .Repeat this
procedure for voltages of 3V ,4V and 5V rms . Copy the results table as shown in figure
, reproduced
at the end of this assignment and tabulate your results . Work out the
magn itude of impe dance of the circuit at 100Hz , calculating the value at ea ch voltage
step and taking an average o f these .
PRACTICAL CONSIDERATIONS AND APPLICATIONS
As well as a
phase shftng
network , the RC circuit can be used as a frequency dependent
circuit as part of a 'thter
A filter is a circuit which will pass some frequencies but attenuate (reject)
others
Consider the circuit in figure .
R
_
0
By the potential divider relationship
2 = V1 (Xc / Z )
/ WC
1
/ COC
R
2
+ ( 1 / 0.)C )
2 r(co CR)2 + 1
now when co CR
1 V2 V1 I
and when w CR > 1,
V2
1 / CO CR
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At low frequencies the capacitor takes little current , so the voltage drop in R is small .
At high frequencies the capacitor short-circuits the output , and most of the input voltage
is dropped across R .When oCR = 1, f = 1/
2TrCR
and at this frequency
V2 = V / 12 ie 0.707 x
Around the frequency a transition occurs between these conditions , as shown in figure
v
i
4-
4 2/7 ck
This type of circuit is known as a low pass filter , as it passes all frequencies below the
frequency known as the cut-off frequency , f
, and attenuates all those
above f c.
Now consider the circuit of figure.
C
p4)
In this case :
V2=V RZ)
V
1
/ R
R2 + ( 1 / co
C )
2
/ ( )CR
r(c.0 CR )2
+ 1
now when co CR (c 1,
V2 ^* COC R V1
and when
O
CR 1,
V2
1
At low frequencies the high reactance of the capacitor restricts the current flowing in R ,
and therefore also the output voltage . At high frequencies the capacitor virtually
connects the output and input directly together . Similarly around the frequency 1/
2nCR
there is a transition between these two states , as shown in figure .
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-jc '
.
This type of circuit is know n as a high -filter , as it passes all frequencies abo ve the cut-
off frequen cy , f , and attenua tes all those b elow f c .
Filters are used extensively in electronics- .High-pass and low -pass types are bo th
com m on , as is a com bination of the tw o types , cal led a ban d-pass f i lter , which passes
all frequencies w ithin a c ertain band , cal led the pass ban d , an d attenuates al l other .
Drawings o f a band pass filter and its respon se are given in figures
CA
O
w t
t
t t 4
-
2 _
_
C
V
l i
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PRACTICAL ASSIGNMENT ON IMPEDANCE OF A SERIES LR CIRCUIT
OB JECTIVE To invest igate the im pedance of series LR circuit to an a.c sinusoidal
waveform ,and see how this varies with frequency .
EXPERIMENTAL PROCEDURE Using a similar technique to that employed in
Assignm ent 21 , we now wish to investigate the impedan ce of a series LR circuit .
Connect up the circuit as shown in the patching diagram of f igure corresponding to the
circuit diagram o f figure .
t y o 44,14
L
Set the frequen cy of the function generator to 800Hz , and ad just the sinusoidal outpu t
of the generator to give 1V rms as given on the m eter .
Take the cu rrent reading for this voltage .
Reset the voltage outpu t to 2V rm s and record th e resultant current .
Repeat this procedure for outputs of 3V , 4V and 5V .
Cop y the results table as show n in figure , reprodu ced at the end of this assignmen t ,
and tabu late your results .
W ork out the average imp edan ce for the LC circuit at this frequency .
PRA CTICAL C ONSIDERATIONS A ND APPLICATIONS Like CR c i rcui ts ,
LR circuits
may be used as filter networks .
Consider the circuit figure .
L-
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With this circuit :
V2 -
i
/R2 X L
2
R
V2
i
/R
+ (27cfL
)
2
Vi
V2
/ R12
This gives a low - pass respon se sim ilar to that in f igure . The ind uctor has a low
impedance to low frequencies (if) , thus the circuit behaves as a shunt
resistan ce at i.f.
and at h igh frequencies (h.f ) the inductor has a high imp edance an d attenuates
h.f
signals .The converse of this is shown in the circuit of figure .
7
V
Here :
L
V2 -
V,
1R2 X
L
2
D L
V2 -
'1R2 + (COL )
2
V2
This is the equation for a high-pass filter .At low frequencies the inductor
shunts the
output to earth , whilst at high frequencies it has a high impedance and the circuit
appears just as a series resistor .The
band pass form of the LR
circuit is
shown in figure
VI
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R
. S
PRACTICAL ASSIGNMENT ON IMPEDANCE OF THE LCR CIRCUIT
OBJECTIVE
To investigate the impedance of a series LCR circuit and compare it with
the impedance's of its constituent components .
EXPERIMENTAL PROCEDURE
Let us investigate a series circuit containing
inductance , capacitance and resistance as in figure .
We know , from kirchhoff ' Voltage Law , that the phasor sum of
VR VL
and Vc w ill equal
V
n
. Connect up the circuit as shown in the patching diagram of figure corresponding to
the circuit diagram of figure .
11(0
I 6
-
epi41
H
F C,
Connect the voltmeter across the input to the circuit ( points marked
P
and S on figure)
and adjust the generator output to give 4V rams output at 500Hz .Record the resultant
current . Transfer your voltmeter to read the voltage across the resistor ( measure
across points P and Q) . Record this voltage . Measure and record the inductor voltage
( points Q and R) and the capacitor voltage (points R and S) .Copy the results table
as
shown in figure , reproduced at the end of this assignment , and tabulate your results .
Draw to scale , a phasor diagram showing Vi
n , VR , VL and Vc
PRACTICAL CONSIDERATIONS AND APPLICATIONS
The LCR network can be
used as a low-pass , a high-pass , or a band-pass network depending on which
component the output voltage is taken across .
Consider the circuit of figure .
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V
1
Here :
X c
. 1 R
+ ( X L
c )
2
1 I coL
V2
R
2
coL -- 1 / (.0C )2
This gives a low-pass response , but above the cut-off frequency both L and C influence
the rate of cut-off . At high frequencies the inductor has a high impedance and the
capacitor has a low impedance , thus the attenuation above f c is at a greater rate than
for the CR or LR filter .
For a high-pass filter the connection is as in figure .
c.
U
2-
This gives a high-pass filter response , as at high frequencies the capacitor is virtually a
short circuit , and the inductive reactance is very high . The circuit acts like a series
resistance at h.f .
The band-pass circuit form is shown in figure .
V
i
R .
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ASSIGNMENT PARALLEL IMPEDANCES
OBJECTIVE
To investigate the impedance of parallel connected components to a
sinusoidal alternating current .
EXPERIMENTAL PROCEDURE
So far we have investigated series circuit of RC ,RL
and found formula for their impedance's , and how these impedance's vary with
frequency . Let us now look at parallel connected impedance's .For series connected
impedance's the reference that is common to all components is that of current , and the
phasor sum of the voltage drops around the loop is zero .For parallel connected
impedance's the common reference is voltage , and the phasor sum of the currents at
any node is zero . This is illustrated in figure .
The voltage v is applied to all of the impedance Z1 , Z2 and
Z3
in common , and
considering instantaneous currents ,
+i
2
+i3 i = o
Set up the circuit as shown in the patching diagram of figure corresponding to the
circuit diagram of figure .
I
Set the function generator to give an output voltage of 4V rms at 1600Hz , as shown on
the meter .
Disconnect link 1 and insert the 0 5mA millimeter .
Record the total current , I . Disconnect the meter 0 5mA meter and replace link 1.
Remove link 2 and connect the meter instead . Record the resistor current , IR .
In similar fashion , measure and record the inductor current , IL , and the capacitor
current ,Ic
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Copy the results table as shown in figure , reproduced at the end of this assignment ,
and tabulate your results .
j
qt
urrehl x
Draw a phasor diagram of the currents IR , IL and lc .Use V as your reference direction .
You can use either the completion of a parallelogram or drawing each phasor from the
tip of the previous one to find the resultant . Figure combines phasor both techniques in
order to show how the reactive current ; if several resistors were connected in parallel
their currents would be numerically added to find the total resistive current . The total
current is than
1 R2
x 2
Where IR is the total resistive current and lx is the total reactive current .
In the circuit of figure the total resistive current is simply IR as marked in the diagram ,
and is given by
IR = V / R
Where R is the resistance , V the voltage .
The reactive current lx is given by
Ix
V/XL+V/Xc=V/X
Where XL = coL is the reactance of the inductor ,
Xc = - 1 / coC is the reactance of the capacitor
And X is the combined reactance of the parallel combination .
Notice that this combined reactance of two parallel reactance's is found by a similar
formula to that used for parallel resistances ,i.e,
/X = 1 /X1 + 1 / X2
This dose not mean that resistances and reactance's can be mixed in the same formula
It dose make it easy to calculate reactance's however , and this is one of the reasons
why capacitive reactance is defined to have a negative value . If it were not so the
formula would be different , and one would have to know whether each of X
1
and X2
were inductive or capacitive
PRACTICAL CONSIDERATIONS AND APPLICATIONS
Circuits which include a
number of parallel branches are often more easily analysed using the admittance
formula rather than the impedance formulae . Calculations should be carried out in
terms of the admittance of the branches and then , if a final impedance value is
required, the reciprocal of the final admittance should be taken .When taking the
reciprocal of a phasor quantity it is necessary to find both the magnitude and the angle
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of the reciprocal . The magnitude of Z
is simply
found as the reciprocal of the
magnitude of Y . The phase angle of Z is minus the phase angle of Y . Figure illustrates
this . V and I are the total circuit voltage and current respectively in the two diagrams
shown , but in one diagram V is the reference phasor , and in the other I is the reference
phase. Since the phase angle is reckoned from the reference to the other phasor the
signs must be different in the two cases .
Working Principle of a Transformer
A Transformer is a static device by means of which electrical power in one ckt. is
transformed into electrical power of the same frequency in another ckt. but with a
corresponding decrease or increase in current / Voltage. The physical basis of a
transformer is a mutual induction between two ckts. linked by a common magnetic flux.
It consists of two inductive coils
which are electrically separated but magnetically linked
through a path of low reluctance. One coil is connected to a source of alternating supply
Voltage, an alternating flux is set up in the laminated core, most of which is linked with
the other coil in which it
produces mutually induced e.m.f. ( Faraday's laws of
Electromagnetism induction e = M dl / dt ). The second coil is closed a current flows in it
and so electric energy is transferred from first coil to second coil. The first coil which fed
supply is called Primary winding and second coil which power drawn out is called
secondary winding.
In brief, Transformer is a device that
a) Transfers electrical power from one ckt to another ckt.
b) It works without any change of frequency.
c)
It accomplishes this by electromagnetic induction.
d)
Where the two electrical ckt are mutual induction influences of each other.
E.M.F. Equation of a Transformer:
Let, N1 = No. of turns in primary winding
N2 = No. of turns in secondary winding
(I) = Maxm. flux in core in webers = B * A
B = Maxm. flux density,
= Area of the core
f = Supply frequency in herz ( H)
Average rate of change of flux = Maxm. flux / 1/4f = 4f wb/s or Volt
Average e.m.f. per turn = 4IVolt
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If flux varies sinusoidal, then r.m.s. value of induced e.m.f. is obtained by multiplying
the average value with
FORM FACTOR.
FORM FACTOR. = r.m.s. value / average value = 1.11
Therefore, the RMS value of the induced e.m.f. in the whole of primary winding equal to
induced e.m.f per turn into the number of primary turn.
Voltage Transformation Ratio
:
El = 4.44 f N Bm A
E2 = 4.44 f N Bm A
From equation No.(i) & (ii) we get,
K ( K is the constant )
'K' is the Voltage Transformation Ratio.
Condition :- a) If K > 1, i.e. N2 > N1, then transformer is called step-up
transformer.
b) If K
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POWER
in
A
C 3 Phase System
Active Power in one Phase = Vp Ip Cosa) Watt
For all the three phase power ( algebraic sum ) = 3 Vp Ip Cosh Watt
Vp =Phase Voltage
L = Line Voltage
Ip = Phase Current
L = Line Current
Three Phase Power = VLIL Cos cD
Watt/ KW
For Star ( Y ) Connection : = Vp = VL / 43 and Ip = IL
All the Three Power =
3 Vp Ip Cos (13 Watt
Three Ph Power =
q3 VL IL Cos (1) Watt
For Delta ( D ) Connection : = Vp = VL and Ip =1L /
All the three Phase Power -4 3 Ph Power =
3 Vp Ip Coscb Watt
3 Ph Power =
q3 VL IL Cos (1) Watt
MEGGERING ( Practical Assignment )
A Megger is mainly used for testing Insulation resistance. Insulation resistance is
the resistance offered to the passage of current across the surface of the insulation. In
any electrical m/c I.R. can be measured between two conductors which are separated
by some insulation or it may be measured between a conducting part and the body /
earth of the machine. Normally all electrical conductors / live parts are insulated
with
rubber, PVC, enamel varnish etc. and the insulating material may deteriorate in course
of time. So checking insulation resistance at regular intervals becomes very important
from the view point of safety and reliability. Normally minimum value of insulation
resistance between two electrically separated conductors or between a conductor and
body / earth should be 1 MO it is less than 1 gp,the electrical equipment under test will
have to be sent for repair/ reconditioning.
Insulation resistance is measured with a megger or insu-tester or motor checker
etc. The unit of measurement is
Mega-ohm ( M41-.)
Objectives
The objectives of meggering are to :
Test & Measure the Insulation resistance between Stator winding & Body/Earth.
Test & Measure the insulation resistance between different windings.
Test & Measure of Stator & Rotor windings and Slip-Ring & Rotor.
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Procedure :
A.
Insulation resistance test between winding body / Earth.
This test is carried out to determine how much insulation resistance is
offered betw een Stator or Rotor wind ing and earth.
Steps -1 Ensure m ain supply is disconnected to the m otor.
Steps -2 Before measu ring insulation resistance, ensure that the body of the m otor is
properly earthed. To do so, connect the L terminal of megg er to the body of the m otor
the E Term inal to earth and rotate the megger han dle. If the reading beco m es Zero,
the earthing is O.K.
Steps -3 Now take Stator terminals ( m ay be m arked A,B , C / Al ,B1,C1-A2, B2, C2 )
Steps-4 Rotate the megger handle at about 160 r.p.m. and note the reading. It
shou ld not be less than 2
m-n ( two Megaohm ).
Steps - 5 Next remo ve the L terminal of the megger and conn ect stator coil individual
( Al / A2 Bl/B2 , Bl/B2 Cl/C2, C1/C2 Al/A2 )
Steps-6 Rotate the megger handle at about 160 r.p.m. and note the reading. It
should be
2M-ohm ( Tw o Megaohm ) or more.
B.
Insulation resistance test between Stator windings and Rotor Ckt.
This test is done for measuring insulation resistance between armature and f ield wd gs.
Steps -1 Take a AC m otor which is not connected to main supply l ine.
Steps -2 Connect the megger L and E terminals to Stator terminal, say A / B /C
and Rotor terminal,
Steps-3 Rotate the megger handle at about 160 rpm. and note the reading. It
should be 2
M -Q ( Two M egaohm ) or more.
C.
Continuity test
This
test is carried out to determ ine the continuity of a particular winding.ln Insu-tester
or m otor checker, low V oltage (9V) range is used for continuity testing. To determ ine
any loose co ntact in the w inding, testing w ith a series test lamp is preferred,
Steps -1 Take a AC mo tor which is not con nected to m ain supply line.
Steps -2 Connect tw o terminals of the con tinuity tester to Stator coil term inals.
If a beep is heard or very low resistance is found then it is o.k.
Steps -3 Connect tw o term inals of the con tinuity tester to Rotor term inals.
If
a beep is
heard or v ery low resistanc e is found then it is o.k.
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F : -
olarization Index ( P. I.
I t is the ratio betw een the I .R. value ( m eggering va lue ) in 60 second s and I .R. value
( meggering value ) in 15 seconds. P.I. is a good appraisal of the condition of the
insulation, clean dry w dgs w ill have a m uch higher index than m oist or dirty ones. P.I. of
less than 1.3 ind icates the need for recond itioning. Index values of 4 or m ore indicate
the insulation to be in very good condition. An index of less than 1 indicates the
developm ent of carbon ized paths through an d around the insulation.
Megger Application Grade
(i)
For Low Voltage
50 Volt Insulation Tester ( Battery Operated )
(ii)
Low Voltage Upto 440 Volt : 500 Volt Insulation Tester ( Megger ) with hand
hold and battery operated.
(iii)
For 1.1 KV
1000 Volt hand hold Megger
(iv) For 3.3 KV
2.5 Killo Volt hand hold Megger
(v)
For 11 KV
5 Killo Volt hand hold Megger
Leakage Current must not exceed 1 / 1000 th (0.0001)mA of Full load Current.
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SAFETY PRECAUTIONS FOR ELECTRICIANS
Introduction
Safety precautions for electricians are the guidance to mold the habit of
electricians to be more responsible to alertness and consciousness to the hazards in
their day to day work.
TYPES OF
SAFETY PRECAUTIONS FOR ELECTRICIAL HAZARDS
1
lectric shocks are easily received and easily be avoided , RISK is not apparent , should
be carefull .
Beware
of live conductors , either bars or insulated .
Never disconnect a plug point by pulling the flexible card .
Always connect live wire through a switch .
5.
Always switch OFF before replacing a blown out fuse.
6.
Never touch an overhead line unless you have made sure that the line is
electrically dead.
7.
Put on safety belt when working above ground on a pole .
8.
Never put a switch ON unless you are sure that all the men are not working in
the line .
9.
When mixing sulfuric acid in water , add acid to the water and not water to the
acid.
10.
When cells are charged in a room maintain good ventilation and never bring
maked light near accumulator (battery).
Use correct size and quality of fuse wire when rewiring blown out fuse wire.
If
a ladder is used it must be hold by another person so that it may slip away .
Do :lot use wire with poor insulation .
Always use portable hand lamp of insulated safety type and provided with a
rubber , plastic or wooden and wire guard .
15.
Safety demands good Earthing . Hence always keep earth connections in good
conditions .
16. Beware working on Motor or other rotating machinerise make that it can not be
set in motion without your permission .
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e w a r e
s t a r t i n g
3 work always follow "F OU R" principles rs.f F IP,Trica Safety .
ISOLATE ( i.
e. SWITCH OFF , FUSE AND PUT
DANGER BOARD .
TEST ( i.e. CHEACK THAT THE CIRCUIT IS DEAD) .
LOCK ( i.e. SECURITY AGAINST UNAUTHORISED SWITCHING) .
SHORT ( i.e. EARTHING OR SHORTING AGAINST ACCIDENTAL CHARGE )
18
isconnect the supply immediately in case of fire rlar electrical apparatus .
19
o not use fire extinguisher on Electrical equipment unless it is clearly marked
an suitable for that purpose use sand or blanket instead .
20 Do not throw water on live Electrical equipment's in case of fire. It is dangerous for
you
21 Do not work on energized circuit without taking extra precautions , Such as the use
of rubber glovan and wooden handless .
22 Do not wear loose clothing , metal watch straps, bangles, finger rings while working
on electrical applicances
23 Before supplying current to electrical, equipmen t , it should be ascertained that the
equipment is properly earthed .
24 Before using portable Electrical things , see that these are well earthed .
25 Do not use a plier as a hammer .
26 Do not put a sharp edged tool in your pocket .
27 Do not use tools like file , knife ,screw driver etc. , without handle otherwise it can
injury the hand .
28
Preach and practice safety at all times. good work
e spoiled by an accident.
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YOURS SAFETY IS IN YOUR HAND
TREATMENT OF ELECTRIC SHOCK FOR ARTIFICAL RESPIRATION
This treatment should be con tinued for at least five hours if necessary .
A person apparently dead m ay be revived by the method (SCH AFER'S) describe below
Rem ove the body from contact with the wire , cable or other conductor .
By
Breaking or Disconn ecting the circuit ,
Dragging the patient away by his coat tails .
The hand being protected by Indian Rubber gloves or
ny dry w oolen m aterials,
such as a cap folded sev eral thickness if possible , wood or any m en cond ucting
m aterial may b e used , If possible without discontinuing the treatment , send for a
doctor .
AFTER REMOVAL
Do not w ait to und er the clothing .
Place the patient on his chest w ith head turned to one side .
Kneel at his side and grasping the lower ribs w ith bo th hand s ,
Gradually throw your w eight on to his body .
Spring quickly back and repeat the mo vem ent fifteen tim e a minute .
Do not leave the patient or stop artificial Resp iration , un til a doc tor arrives
Keep the patient warn .
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